Question

19. Hot magnesium metal reacts with nitrogen gas to produce magnesium nitride:3 Mg(s)+ N2(g)→Mg3N2(s)What mass of N2gas is required to completely react 1.32 g of magnesium metal?a. 0.507 gb. 0.0543 gc. 0.0181 gd. 0.254 g

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Answer 1

(19) Given reaction is

3Mg(s) + N2(g) ------------> Mg3N2(s)

Given mass of Mg=1.32 g and molar mass=24.305 g/mol.

Moles of Mg=mass/molar mass=1.32 g/24.305 g/mol=0.0543 mol.

From the balanced equation, the moles of N2 required=(1/3) x moles of Mg=(1/3) x 0.0543 mol

Moles of N2=0.0181 mol.

Molar mass of N2=28.0134 g/mol.

Mass of N2 required=moles x molar mass=0.0181 mol x 28.0134 g/mol=0.507 g.

Therefore option (a) is correct.

Please let me know if you have any doubt. Thanks.